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G Load during payload drop
Some days ago I saw the accident of the airtractor that, just after a drop, suffered a spar failure in both the wings. Luckily the pilot survived.

If load factor in his simplest formula is:
n=lift/weight
So if an aircraft experiences a sudden loss in weight, is natural to think that the aircraft will experience a temporary increase in G load during the maneuver.
My doubt is about where that forces are going to be applied.
Let's now suppose that during the complete maneuver, the aircraft is going to maintain straight and level flight without losing or gaining a feet of altitude. In this case, will the complete aircraft be under that load (like during a turn), or the load is going to be applied only on the wing(s).
Thanks in advance for the answers!

If the load factor > 1 then the aircraft is not in straight and level flight.

As stated in the Consolidation and Analysis of
Loading Data in Firefighting Operations of the FAA:  The effect of relatively large aircraft weight changes due to the release of retardant causes uncertainty to the assessment of ongoing structural integrity, since mission weights have not been routinely tracked.

Preliminary evidence obtained from some recently instrumented aircraft suggests that the high g levels occur both at high gross weight prior to the retardant drop and at lesser gross weights after the retardant drop. The overall rate of damage accrual determined from measured acceleration data should account for representative changes of aircraft weight in a realistic or conservative manner. https://www.tc.faa.gov/its/worldpac/techrpt/ar05-35.pdf

Your skepticism is justified, and it's not just a question of how significant the change in G-load created by dropping a payload would be.  Dropping weight from the fuselage creates no increase in the stress on the wing-fuselage junctions.  In fact, dropping payload from the fuselage decreases the stress on the wing-fuselage junctions.  (Dropping weight from the wings, however, would increase the stress on the wing-fuselage junctions.)
Assume an aircraft weighs 10,0000 pounds without payload, and can carry a 10,000 pound payload.  Assume the weight of wings is negligible, so the fuselage weighs 10,000 pounds without payload.  Assume the payload is carried in the fuselage.
In straight-and-level flight with no payload, 10,000 pounds of force must be exerted on the fuselage by the wings.  So 5000 pounds of force must be transferred through each wing-fuselage junction.
In straight and level flight with payload, 20,000 pounds of force must be exerted on the fuselage by the wings.  So 10,000 pounds of force must be transferred through each wing-fuselage junction.
Assume the plane is in straight-and-level flight with payload, and then the payload is dropped instantly.  The weight of the aircraft is cut in half.  The airspeed and angle-of-attack of the wings will initially stay the same, so each wing is still generating 10,000 pounds of force, all of which is transferred through the wing-fuselage junction.  There is no increase on the stress on the wing-fuselage junction.  The aircraft is now in a +2 G condition and will accelerate upwards.  I.e. the flight path will curve upwards.  This upward acceleration will create stress on the mounting points of heavy components such as the engine, battery, pilot's seat (including pilot), etc, but will not increase stress on the wing-fuselage junction.1
The upward acceleration will have some tendency to decrease the angle-of-attack of the wings, decreasing the total lift force, and decreasing the total amount of force transferred through the wing-fuselage junctions, and decreasing the total G-loading, but this will be offset by the plane's inherent pitch stability dynamics (assuming that the payload was carried at the CG), which will tend to pitch the nose up to conform to the upward curve in the flight path and restore the original angle-of-attack.
At any rate, we can see that there is no tendency for the wing-fuselage junctions to fail in a way that would cause the wings to rise upwards relative to the fuselage.  The same amount of force is being transferred from the wings to the fuselage as was the case before the drop, but that force is now being used to accelerate the fuselage upwards rather than to support the payload.
Now assume that the weight of the wings is not negligible.  Assume the wings weigh 500 pounds each and the fuselage weighs 9,000 pounds without payload.  Now in straight-and-level flight with a 10,000 pound payload in the fuselage, 19,000 pounds of force must be exerted on the fuselage by the wings.  So 9,500 pounds of force must be transferred through each wing-fuselage junction.
Now when the payload is dropped instantly from straight-and-level flight, and the wings continue to fly at the same airspeed and angle-of-attack and generate 10,000 pounds of lift each, the aircraft will be in a 2-G condition and will accelerate upwards.  1000 pounds of lift force will be "absorbed" by each wing, and 9000 pounds of lift force will be transferred through each of the two wing-fuselage junctions to the fuselage.  So dropping the payload from the fuselage has actually decreased the stress on the wing-fuselage junctions, because some of the lift force from the wings is being used to accelerate the wings upwards, rather than to support the payload in the fuselage!
Now let's change the picture and assume that the payload is carried in the wings, not the fuselage.  Let's assume the wings weigh 500 pounds each empty, and 5,500 pounds each loaded.  Let's assume the fuselage weighs 9,000 pounds.  In straight-and-level flight in the loaded condition, the wings are generating 20,000 pounds of lift total, of which 9,000 pounds are being used to lift the fuselage.  4,500 pounds of force must be transferred through each of the two wing-fuselage junctions.
Again, assume the plane is in straight-and-level flight with payload, and then the payload is dropped instantly.  The weight of the aircraft is cut in half.  The airspeed and angle-of-attack of the wings will initially stay the same, so each wing is still generating 10,000 pounds of force, so the aircraft is accelerating upwards in a 2G condition.  Of the 10,000 pounds of force being generated by each wing, 1000 pounds is "absorbed by" the wing and 9000 pounds must be transferred through each of the two wing-fuselage junction to the fuselage.  The force on each of the wing-fuselage junctions has doubled.
Take-home lessons--

Dropping weight from the fuselage does not increase the stress on the wing-fuselage junctions.  Rather, if the weight of the wings is not negligible, then dropping weight from the fuselage decreases the stress on the wing-fuselage junctions as the aircraft accelerates upwards.

Dropping weight from the wings does increase the stress on the wing-fuselage junctions as the aircraft accelerates upwards.

You also asked--

Let's now suppose that during the complete maneuver, the aircraft is
going to maintain straight and level flight without losing or gaining
a feet of altitude.

This would require the pilot to shove the stick forward to reduce the angle-of-attack, and lift force generated by the wings, at the instant the payload was released.  ("Shove" only applies to our thought experiment where the entire payload is released instantly; if the payload is released more gradually then a more gradual forward movement of the stick would be required.)  Reducing the angle-of-attack of the wings doesn't create any unusual loads on the aircraft.  And the aircraft would remain in a 1-G condition.
Footnotes--

Note that for a given total lift force generated by the wings, the higher the total mass and weight of the aircraft, the lower the G-loading, and thus the lower the stress on the mounting points of heavy objects of fixed weight such as the motor, battery, occupants' seats (including occupants), etc.  That's why for many aircraft, the published maneuvering speed is lower when the total mass and weight of the aircraft is lowered.  Lowering the allowable maneuvering speed also lowers the total amount of force that can be generated when the wing reaches the stall angle-of-attack while flying at the maneuvering speed, which keeps the maximum possible total G-loading constant, which keeps the maximum possible stress exerted on the mounting points of heavy objects of fixed weight constant.  (See related ASE question Why does maneuvering speed vary with weight?).  However, this approach doesn't make sense if 1) the critical points of concern that will fail first are the wing-fuselage junctions and 2) any added weight is added to the fuselage, not the wings.  In this case, as this answer shows, as long as the weight of the wings is non-trivial, then to keep the stress exerted on the wing-fuselage junctions constant, the maximum allowable lift force should be decreased as the fuselage loading is increased, which means that the maneuvering speed should be decreased as the fuselage loading is increased.  The opposite is true if the extra weight is being added to the wings rather than the fuselage.  See also this related ASE answer -- How does an aircraft's weight affect the V-n diagram?.

Thanks a lot for the detailed answer! So it's correct to think that, in the first case, there is a negative load on the wing-fueselage sections. About the physical sensations, in cockpit or onboard, that are not generally felt, like Micheal wrote in his post, is because the payload that is drop is generally a "small" fraction of the Total weight, so realistically speaking the variation is probably around +-0.2g. Your answer matches also about the existence of the MZFW (especially applied on the second case). Thanks!

There are a couple of misconceptions in your question that I believe are leading you to a false conclusion that G-loads from releasing payload may have literally ripped the wings off this airplane:

Spraying liquid will not cause a "sudden" loss of weight.  While fire fighting aircraft are capable of releasing large quantities of liquid all at once, agricultural sprayers like the one depicted are designed to disperse over a much larger area.  Hence the gross weight lost on each pass is pretty negligible.

Losing gross weight will not cause a sudden and appreciable increase in load factor.  Whatever the math might say, I can tell you from the personal experience of standing in the cargo bay of C-130s as thousands of pounds have rolled out the back end, that there is very little perceptible change in felt G forces.  (Conversely, I have been walking forward in the cargo bay when the pilot put on a roughly 2G turn and my knees buckled, sending me to the deck and breaking a tooth...)

So, while load factor will change somewhat when weight is changed suddenly, it would be well within structural limits that the aircraft is designed for, and would in no way account for the type of dramatic spar failure as depicted in the photo.
If you are interested in a similar incident, a C-130 tanker tragically and dramatically had its wings snap off in a similar manner.  More info in the link here:  C-130 crash
I don't recall all the details of this, or the incident you describe, but generally there is corrosion, and/or fatigue cracking, and when the catastrophic failure finally occurs it is as a result of the pilot applying G loads during the pullout, not because of the gross weight change from dropping payload.

Thanks for your answer Michael! It helps a bit to solve my doubt; probably most of the load is going to be applied only on the wings/torsion box of the aircraft. The air tractor in the picture was not damaged during spraying but during a complete release of water, you can find the video here: https://www.instagram.com/p/CXWzus1FyEW/ The aircraft structure was probably compromised well before the accident.

@Spitfire01, thanks for the video link!  You can definitely see that the pilot starts a pretty good pull just before the wings snap.  Amazing that he survived...

Just dumping load increases the G force you feel inside the plane because the same amount of lift produced by the wings.  What changes is the amount of downward force gravity exerts on the plane + stuff inside it, because there's less stuff.  So if speed and pitch stayed the same, you'd expect the wing load in terms of force in Newtons to stay nearly constant, even though people in the plane briefly feel an upward acceleration.  Changing the weight distribution probably will lead to some pitch change, though, and of course the pilot pulling up will be a factor.  (@Spitfire01)

TL:DR: my point is that even if you did feel an acceleration while standing in the back as the load drops, that doesn't mean there's more force on the wings.  Just that the same lift is now opposed by less weight.

Related to quantity you're right in general indeed, but this case is different: "pesticide application was canceled due to bad weather conditions. The pilot discharged water from the tank onto the landing site and began the climb and his wings broke and the aircraft hit the ground.".  The tank was emptied at once. Maybe this was not so much water, we don't know.


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Wouldn't a triplane be much more efficient than a monoplane of same wing area and span?
Note: not a duplicate of Can biplane or triplane designs be revived with modern materials?
This question compares a high performance triplane with high aspect ratios and high gap, to a monoplane. It does not explore materials engineering.
Wouldn't a triplane be much more efficient than a monoplane of same wing area and span?
Take a small GA plane, 1,000 lbs, 100sq ft of Wing area, 25' span, 4' chord, a pretty typical LSA/microlight. So, an aspect ratio of roughly 6, and a wing loading of 10psf.
NOTE:  With a gap of 2 chord, I understand the efficiency of multi wing aircraft is almost the same as a monoplane. see graph.
Now make it a triplane, which reduces the chord to roughly 1.3, increasing the aspect ratio from 6 to 18, just like a glider.  That's a 300% increase in efficiency, no?
Instead of a heavy spar for a thin wing, you could just put the stabilator at the wing tips joining the 3 wings: so no extra wing strut interference drag ( expect for a small amount of extra interference drag at 2 wings roots).
So:
less weight( same wing area, but smaller lighter wings),
smaller engine needed ( so could even go with smaller wing area)
less induced drag (due to 3 times the aspect ratio)
a less form drag ( same span, but thinner wings (66% THINNER(??) due to 66% smaller chord)
a little more wing root interference drag,
no extra wing strut interference drag
less induced drag due to 1/3 the wing loading
NOTE: All this could only happen with a gap of 2 chord or more, where I'm told 2 wings act independantly without interfering with each other, just like a main wing and horizontal tail( which is a biplane of sorts)
Is this correct, at least conceptually?

Related: https://aviation.stackexchange.com/questions/21845/can-biplane-or-triplane-designs-be-revived-with-modern-materials

You said "see graph" but I don't see anything, did you mean to include one?

A mono would have a span of about 40 feet and a chord of 2.5, way to wide for hangers and taxi ways at most flying clubs. Also, the spar would be pretty heavy I think.

I'm voting to reopen this. It looks to me like the asker's confusion is caused by one specific misconception that the answers to the linked question don't address. So we need new answers in order to address that.

@Fred feet or meters? 40 feet is fairly generic for (full scale) GA aircraft.  Slats and flaps function as "2nd and 3rd wing", compare the Fiesler Storch to biplanes.  When you retract them, like airliners, you get a huge speed envelope that bipes cannot match given equal power.  But for a simple, reliable GA aircraft, I'm honestly surprised more aren't around.  For very high aspect, check out powered gliders.

No. Even ignoring additional drag from wing interference, there's little or nothing to be gained by going to a triplane configuration. In the case of a vehicle with multiple wings, the effective aspect ratio of the vehicle is different than the aspect ratio of a single wing.
The induced drag of the vehicle strongly depends on span; aspect ratio is just a convenient dimensionless parameter to represent this.  , If you keep the same wing area and don't change the span, the induced drag will not change significantly.  This is the case with the triplane you describe.
For some intuition on why this is the case, remember that induced drag is an effect of downwash caused by wingtip vorticies, the strength of which is related to the total amount of lift the vehicle produces.  This is related mainly to the vehicle wing loading $W/S$.   Vehicles with large spans reduce this effect by moving the tip vorticies out further from the center of the vehicle.
In the case where you have the smaller wings, the induced drag of each is reduced by ~1/3 but you need to sum the drag from each wing into the total vehicle drag.
$C_D = C_{D_0} + C_{D_{i,1}}+ C_{D_{i,2}}+ C_{D_{i,3}} + C_{D_\mathrm{interference}}$
You can see this mathematically by considering the dimensional form of the equation for induced drag $D_i$.
$D_i = qS\frac{C_L^2}{\pi e \mathrm{AR}}$
In level flight, $C_L = \frac{W}{qS}$ and recalling $AR = \frac{b^2}{S}$,
$D_i = \frac{1}{\pi e}\frac{1}{q}\frac{W^2}{b^2}$
This shows it's really the span loading, not the aspect ratio, that drives induced drag. Aspect ratio is just convenient shorthand.

If you keep the same overall span width, keeping the same wing area, you are reducing the chord by 1/3, and you are increasing the effective span by 300%, so your span loading goes down by 66%.  Frontal area is unchanged as the the wings are 1/3 as thick, therefore form drag is unchanged,   At a gap and stagger  of 2 chord,  I'm told there is no interference drag, just like a main wing and tail on a conventional plane, which is really a biplane of sorts anyways.

The span loading of each wing goes down by 66%. But you're tripling the number of wings.  So the span loading of the vehicle as a whole doesn't change.

So what you're saying: If I have an airplane, 30 span, 6' chord, and now make it a high performance glider with a 90' span and 2' chord, it will be way more efficient. Now if I cut up the glider wing and make it a triplane, it will no longer be a high performance glider, it will have the same drag as the original 30 span, 6' chord?

Yep exactly; that's why high performance gliders have one long thin wing and not several short thin wings.

@Fred another angle to consider: three sets of wings will have three sets of wingtip vortices. Consider the spanwise lift distribution, it makes no sense to have multiple inefficient wingtips and wing-body interference zones; a single wing works better.

Now make it a triplane, which reduces the chord to roughly 1.3, increasing the aspect ratio from 6 to 18, just like a glider. That's a 300% increase in efficiency, no?

Not really.
First, while reduced chord does reduce friction drag, the interference between the three wing surfaces increases it again. It's the span that matters more than the aspect ratio.
Second, you've just tripled your wings' frontal cross-section. If your purpose is merely to stay aloft with minimal effort, that's not a big deal. If you're trying to get somewhere, though, the greatly increased parasitic drag will matter, more so the faster you fly.
Simply put, a triplane design is better for minimum gallons per hour and a monoplane design for maximum miles per gallon.

I'm told there is minimal interference with a gap of 2 chord.  Frontal area hasn't increased at all actually. The depth of a 12% foil at 4' chord is 5.76". The depth of a 12% foil with 1.3' chord is 1.87" x 3 = 5.67".  Actually the frontal area has decreased a bit.

Three wings will give you at least three times as much drag as one wing, won't it? So the "300% increase in efficiency" is more like a 0% increase in efficiency at best, and a large decrease in efficiency at worst. Would you like to address this in your answer?

Reduced chord does exactly nothing to induced drag, unless the wing becomes lighter, so less lift is required.

Wouldn't a triplane be much more efficient than a monoplane of same
  wing area?

In strictly forward flight probably, but what are you looking to gain?
You could make the mono wing have a higher aspect ratio and gain the same efficiency, all while being able to use a normal tail stabilizer and predictable flight dynamics.
There's only 2 disadvantages going this route. First the aircraft won't have as fast of a roll response because of the long wing span. Second the long wingspan may be harder to store depending on your ground operations.
But maybe you want to build an efficient aircraft that could land on public roads. So you need the wing span as short as possible while still being efficient, would your idea work?
Probably, but not without it's drawbacks. First how will you pitch stabilize it? If you use a conventional tail, you need to take your "2 cord" rule into effect for this "4th" wing. (I'm assuming for now that the 2 cord distance is correct, but I'll put my own thoughts in on this later).
Mono planes generate lift from 1 spot, pretty easy to predict where (where the wing is) and placing that wing just behind the center of gravity means the stabilizer tail doesn't have to produce much down force to keep the plane in stable flight.

However your center of lift will be moving depending if your the wind hits one of your wings in a different way than the others. You will also have to consider how you're going to make the lift of all 3 wings equal roughly the center of gravity.
Anything off of this mark and would have to provide enough down force to compensate, reducing efficiency.
Now my thoughts on if 2 cords apart is enough for a wing. It probably is in forward flight, but fluid dynamics are hard to predict and when that aircraft starts pitching up and down and the vortexes each wing generate get aimed in different directions your wings will interfere with each-other.
Take a look at the quickie aircraft. It was a "bi-plane" design similar to what you propose. Notice how the wings are not only far apart, but also on separate vertical levels. This is to help dampen pitch response, and give the wings more breathing room in non level flight.
Using this idea for a 3 wing aircraft seems even more difficult, as the more vertical distance your wings take up the less pitch authority you have. A little bit of dampening is okay, but too much and you can't pitch the aircraft.
You'll also notice the quickie doesn't use a normal tail stabilizer (because of the problems discussed above). So differing lift coefficients between the wings are used, but that does make the aircraft harder to fly in general.

Anyway if you're really interested in this idea I'd recommend checking out the quickie project and learning what challenges/breakthroughs they faced/made.

This is a good answer, but the concept of "down force" on a horizontal stabilizer belies its true function, to set pitch of wing.  This is why the power conscious Piper Cub uses a low drag thin flat plate. At proper AOA, there is neither up or down force on the "stabilator".  Only a little elevator trim down force is needed to compensate for CG/Wing Clift offset needed to set up positive static stability.  Not 250 lbs!!!

@RobertDiGiovanni Thank you for the clarification, I was meaning to say that in a normal mono wing aircraft the horizontal stabilizer doesn't need to do much, where in this configuration it may need to do much more. Is that correct? If so I will edit my answer to make it more clear.

@RobertDiGiovanni Oh I see you got 250 lbs from the picture. Sorry I just found it on google images to reference the dynamics involved, didn't pay attention to the numbers it used. I'll find a picture without numbers

I would try 2×18:1.5 bipe first.

@YAHsaves: Distance is not gap. Gap is the vertical distance between 2 wings, stagger is the horizontal distance between 2 wings. I would uses a gap of 2 chord, and a stagger of 0, so the CG is the same, whether it's 1,2 or 3 wings.  Triplanes have been around for 100+ years. They use the same tail configuration as a monoplane.

I know, this answer is almost three years late, but the existing answers leave important issues open.

Now make it a triplane, which reduces the chord to roughly 1/3, increasing the aspect ratio from 6 to 18, just like a glider. That's a 300% increase in efficiency, no?

No. Reducing the chord will do exactly nothing to induced drag. With the same span and the same mass, induced drag remains unaffected. However, by vertically spacing the wings generously, a slight reduction of induced drag will be possible because more air is involved in lift creation.
But only if mass remains unchanged.

Instead of a heavy spar for a thin wing, you could just put the stabilator at the wing tips joining the 3 wings: so no extra wing strut interference drag ( expect for a small amount of extra interference drag at 2 wings roots).

Again, no. If you have no shear transmission between the single wings, each has to carry its lift just like any cantilever wing. With 1/3 the chord and 1/3 the lift per wing, this will require three massive spars because each spar cap will be three times larger than one of a scaled down cantilever wing.
Bi- and triplane wings are lighter for two reasons:

Lower maximum speed of the whole airplane, so the maximum dynamic pressure for which the wings need to be sized is much smaller, and
Bracing allows the lower wing to act as the tension member and the upper wing as the compression member of a truss. But that requires shear transfer between the two, which is traditionally done by bracing.

Without bracing, the triplane wing will become much heavier than the monoplane wing of the same span. Joining the tips of the three wings will not change this - at the wingtip is no bending moment, so there is no benefit in transferring loads between the tips.
Bracing will only help when done along the span, not at the tip alone. And the viscous drag of bracing will require a much larger and heavier engine to keep this triplane in the air. Sure, now the braced wing is lighter, but those gains will be more than eaten up by the heavier engine and higher fuel load for a meaningful range.

Also three years late...
From this comparison of mono, biplane and triplane efficiency, in level flight the triplane is only 70% as efficient as the mono, and the biplane is slightly better at 74%. The penalty lessens with increasing load.

Efficiency is lost as high pressure appears on the underside of the upper wing which negatively affects the low pressure on the topside of the lower wing. Staggering the upper wing ahead of the lower can reduce the penalty by 5%, but efficiency is still way lower than the monoplane.

Actually your data indicates that wing tip and fuselage effects are a greater percentage of drag at low AoA, whereas multiple wings "catch up" at higher AoA.  But this is much better than the baloney that a biplane will only make 20% more lift than a monoplane.

Your approach indeed will work for very low cruising speed, providing more efficient lift with multiple higher aspect wings. This is exactly what soaring eagles, condors, and vultures have on their wing tips. These work great for slow circling flight while hunting an area.
However, these wings are speed limited, rapidly producing too much drag at higher speeds.
The albatross uses more speed and lower drag to generate adequate lift and migrate 1000s of miles.  Notice the difference in their wings.  The entire wing is longer and thinner, without multiple "winglets".
Study of these birds parallels evolution in aircraft design.  Engine efficiency improvements and higher speeds at higher altitudes (also an advantage) has lead to the lowest drag mono-wing designs being most common.
But for walking speed free flight models, built light but strong, a thin undercambered very high aspect biplane is a delight to fly.  They just have problems making any forward progress in a headwind.

Now make it a triplane, which reduces the chord to roughly 1.3, increasing the aspect ratio from 6 to 18, just like a glider. That's a 300% increase in efficiency, no?

No, because you now have three times as many wing tips.
Wing tip losses occur when high pressure air flows from the underside of the wing, around the tip to the low pressure area on the top. Increasing aspect ratio helps by reducing the amount of wing near the tip, but if you reduce the chord and increase the number of wings, you still have the same amount of wing area 'near' a tip and subject to tip losses.

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Why does a force on a propeller act 90° "ahead" due to gyroscopic precession?
This is a common explanation:

A tailwheel airplane is on the takeoff roll and the tail comes up, the
propeller disc rotates forward. That's akin to someone pushing on the
top of the disc. The force results 90 degrees ahead, or on the right
side, pushing the nose left.

I do not understand the term "90 degrees ahead".
In my mind, in this example, we are viewing the propeller spinning clockwise.  If a force is applied at the top (12 oclock), then the resulting gyroscopic force on the right (3 oclock) should be called 90 degrees after.  If the force is applied at 12 oclock, spines 90 degrees and then a resulting force at 3 oclock, why is that called before instead of after?

Seeing as the statement you cited actually made no reference at all to the word "before", there may be grounds for closing this question as in need of clarification.

This is just a semantic discussion on what "ahead" and "after" means in the context.  If you're describing it as a future action/result, "blade moves ahead 90 degrees" from the vertical is one way of looking at it, and that's how I would interpret the block quote.  If you're describing it as a past event, you might say the blade moved to a position 90 degrees after the vertical.  In either case, the action started at 12 o'clock provides the resultant force at 3 o'clock; ahead, behind, whatever.

Another problem with this question is that the title appears to be asking a question about the mechanisms of the physics (which might take a PhD-level understanding to answer, or at least a lot of math), while the body makes it clear that it is actually just a question about the meaning of words.  I'll try to help by adding quotes around "ahead", but I suspect the question will be soon closed.

If I say "my brother was born ahead of me", you might reasonably assume that my brother is older than me.  That his birth took place at an earlier point in time.
You seem to be assuming that "ahead" refers to the position that any given blade of the propeller occupied at an earlier point of time.  I.e., a position displaced opposite to the direction that the propeller is rotating.
But if you are driving a car and your passenger says "the light's green, go ahead", would you take that a suggestion to go backwards?  Or to go forwards?
The meaning of "ahead" in the passage quoted in the original question is correct and reasonably unambiguous-- it means forwards along the direction that the propeller is rotating.
Note that the original passage did not contain the word "before".

Thank Quiet Flyer, I believe you interpreted my misunderstanding correctly and have given me the context to reconcile the description with my understanding.

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What are E2 and E4 airspace?
We have a number of questions here about E2 vs. E4 airspace (see summary of links in this answer) but what does that mean? As far as I can see, neither the AIM nor the ATC Orders mention E2 or E4 airspace, so what are they? What are the practical differences for pilots and controllers?

Deleting some comments-- still want to point out that if this question really is meant to include a question about the practical effects of the E2 / E4 distinction for pilots and controllers, then it is a very broad question.  Not necessarily a bad thing.

“What are the practical differences for pilots and controllers?”  Yes, please help me understand why it matters... to more than one person here.

With some comments deleted, the two comments immediately above seem somewhat out of context-- really as of now it seems they apply more to this question https://aviation.stackexchange.com/questions/71406/in-the-us-how-does-the-distinction-between-e2-and-e3-e4-airspace-affect-pilots .

According to this website, which directly quotes an internal memorandum of the FAA, the class E airspace can be categorized as below:

E1 – Class E Airspace at and above 14,500 feet MSL
E2 – Class E
airspace areas designated as a surface area for an airport
E3 – Class
E Airspace Areas Designated As An Extension To A Class C Surface Area
E4 – Class E Airspace Areas Designated as an Extension to a Class D
or Class E Surface Area
E5 – Class E Airspace Areas Extending Upward
From 700 Feet Or More Above The Surface of The Earth
E6 – En route
Domestic Airspace Areas
E7 – Offshore Airspace Areas
E8 – Class E
Airspace Areas Designated As Federal Airways

The memorandum is also available as PDF.
Apart from this reference to the FAA internal memo, I could not find any official FAA documents explaining the categories.
For pilots and controllers there seems to be not many differences, but for operators of small unmanned aircraft (i.e. drones) there are.

Unfortunately the phrase "Designated as a Surface Area for an Airport" is somewhat confusing and would seem at first glance to also include E4 airspace, until we read further down the list.  Also a reader not already well versed in the language at play here would tend to assume the E4 airspace does not touch the surface.  To make everything more complicated still, the FAA often uses the simple term "Surface Area" even without the "designated for an airport" qualifier to mean the same thing as a "Surface Area designated for an airport", i.e. including E2 but not E4 airspace-- see for example

(continuing) see for example the Pilot-Controller Glossary (page S-8) (https://www.faa.gov/air_traffic/publications/media/pcg_10-12-17.pdf) , as well as this document noting an airspace change at KACV https://www.federalregister.gov/documents/2017/07/07/2017-14219/amendment-of-class-e-airspace-arcata-ca-fortuna-ca-and-establishment-of-class-e-airspace-arcata-ca .  Still, good concise answer.

In brief-- E2 airspace and E4 airspace are both examples of Class E airspace that extend all the way down to the surface.  A dashed magenta line that actually encircles the airport for which it is designated normally depicts E2 airspace.  A dashed magenta line that does not actually encircle the airport for which it is designated normally depicts E4 airspace, which is called an "extension".  Both E2 and E4 airspaces serve to bring the Class E visibility and cloud clearance requirements all the way down to the surface.  But some FARs may be construed as treating E2 airspace differently than E4 airspace.  See the links given at the end of this answer for more on that.
Near airports with part-time control towers, airspace sometimes has a different designation at different times of day-- specifically, the designation may change when the control tower at the associated airport closes or opens.  When the tower closes, Class D airspace may change to E2 or Class G airspace.  If the Class D airspace changes to G, then so will any associated extensions.  If the Class D airspace changes to E2, any associated E4 extensions may either be defined to stay E4 or to change to become part of the E2-- it varies, and there is no way to tell which is the case from the sectional chart or even the Chart Supplement.  The fact that very few pilots are aware of this latter point suggests that it rarely matters in actual practice.
Links to specific examples of E2 and E4 airspace on the VFR sectional charts are given in the related ASE answer In the US, how does the distinction between E2 and E3/E4 airspace affect pilots?.

Now in more detail--
"E1", "E2", "E3", "E4", "E5", and "E6" are just shorthand for different airspace descriptions, all of which apply to Class E airspace.  "E2" and "E4" and the rest cannot be found in the FAR or the AIM, but they are  used in other FAA documents.
The ultimate source for these descriptions is the FAA's "Airspace Designations and Reporting Points" document, current edition FAA Joint Order 7400.11E, effective September 14, 2020.
"E2" airspace means all the airspace descriptions that begin with "E2" on this document, found on pp E-1 through E-142.  The heading for this section is "6002. Class E Airspace Areas Designated as Surface Areas.  The Class E airspace areas listed below are designated as a surface area for an airport."
"E4" airspace means all the airspace descriptions that begin with "E4" on this document, found on pp E-157 through E-225.  The heading for this section is "6004. Class E Airspace Areas Designated as an Extension to a Class D or Class E Surface Area.  The Class E airspace areas listed below consist of airspace extending upward from the surface designated as an extension to a Class D or Class E surface area."
Since the FARs do not reference "E2" and "E4" directly, the phrases in the titles and descriptions above are the key to understanding which FARs apply to E2 airspace and which FARs apply to E4 airspace.
In particular, the language of the "Airspace Designations and Reporting Points" document arguably signals that when we come across the phrase "designated for a airport" or "designated to the surface for an airport", then if we are talking about Class E airspace, we must be talking about E2 airspace, not E3 or E4 airspace.
We can also note the following--

Both E2 and E4 airspaces are normally depicted by a dashed magenta border on a sectional chart.

Sometimes a Class D airspace (and also the core "surface area" of a Class C airspace) can be designated to change to E2 airspace when the tower closes, in which case there will be no specific depiction of the E2 airspace on the VFR sectional chart.

A body of E2 airspace always includes the airspace directly over the airport whose approaches are being protected.

A body of E4 airspace never includes the airspace directly over the airport whose approaches are being protected.  Rather, the E4 airspace abuts some other airspace, such as E2 or Class D, that includes the airspace directly over the airport whose approaches are being protected.  That's why E4 airspace is called an "extension".

Admittedly the terminology applied to E2 and E4 airspace in the "Airspace Designations and Reporting Points" document is rather awkward.  Considering that E4 airspace also goes to the surface, it's unfortunate that someone didn't come up with a better term than "Designated as Surface Areas" or "Designated as a Surface Area for an airport" to describe the E2 airspace.
Many examples can be found where the FAA is using the phrase "surface area" in a context that is clearly meant to exclude the surface-level Class E "extensions" (E3/E3a/E4 airspaces).  On the other hand, air traffic controllers often use the phrase "surface area" to mean the entire column of airspace that extends up from the surface in any given location, including any surface-level Class E "extensions" that may be present.  There are many cases in various FAA Orders where the word "surface area" is used in a way such that it is not clear which of these two usages is intended.
Both E2 and E4 airspace evolved from the earlier "Control Zones".  To read more about this, see this related ASE answer.
Perhaps someday the notations E2, E4, etc will be inserted directly into the FARs and the AIM.  That would greatly help to clarify the meaning of certain FARs.
A note on the AIM -- the Airman's Information Manual is not a regulatory document, but ideally should shed a clear light on the meaning of the regulations.  The phrases used in the AIM to refer to E2 and E4 airspace are similar to, but slightly different from, the phrases used in the "Airspace Designations and Reporting Points" document.  In general the language and meaning of the AIM is congruent with the language and meaning of "Airspace Designations and Reporting Points" document as outlined in this answer.
Related ASE questions and answers:
Q: In the US, how does the distinction between E2 and E3/E4 airspace affect pilots?
A: In the US, how does the distinction between E2 and E3/E4 airspace affect pilots?

It's still there, I just removed the word "handy".  Maybe someone else can do a better job of making a truly succinct answer but the question as a whole is a broad one.  Hopefully the first few paragraphs of my answer are enlightening . I encourage anyone w/ an interest to borrow freely from my answer if they want to attempt to formulate something more concise.

Class C airports also revert to Class E at night, presumably E2.

@randomhead-- yes, you are absolutely right; edited.  Thanks.

So when I am transitioning from E2 to E4, (or vice versa) what must I do?

@MichaelHall-- well, what are you flying?  In the vice versa case, if you are flying an ultralight, a case can be made that you are legal in the E4 but not in the E2.  If you are flying under a SVFR clearance and you cannot maintain standard VFR minimums for Class E airspace, a case may be made that you are legal in the E2 but not in the E4.  But, alternative cases may also be made.  So... your mileage may vary.  If you are in contact with in a controller ,e.g. when you requested a SVFR clearance, ask them if it applies to the "extensions" (E4 airspace) or not.

E2 = dashed magenta border that actually surrounds the airport whose approaches are being protected = "6002. Class E Airspace Areas Designated as Surface Areas. The Class E airspace areas listed below are designated as a surface area for an airport."
E4 = dashed magenta border that does not actually surround the airport whose approaches are being protected but rather adjoins some other airspace that surrounds that airport = "6004. Class E Airspace Areas Designated as an Extension to a Class D or Class E Surface Area. The Class E airspace areas listed below consist of airspace extending upward from the surface designated as an extension to a Class D or Class E surface area.
Source of quotes -- the FAA's "Airspace Designations and Reporting Points" document, current edition FAA Joint Order 7400.11E, effective September 14, 2020.  All the entries in the relevant sections start with "E2", or "E4", as appropriate.

Second try, for brevity

This is the best answer I have seen on this fascination of yours in years!  It gets my upvote.


text:
Are there any/many US airports (with instrument approaches) still without RNAV approaches?
GPS approaches being so trivially easy to establish and maintain, are there any/many US airports with instrument approaches which do not have at least one RNAV approach?

According to the May 2022 TERPS catalog.
FAA Digital Procedures Website
There are over 3,000 airports in the U.S. national airspace with a RNAV approach comprising over 7,000 RNAV (GPS), RNAV (RNP) and GPS approaches.
There are 33 airports without any RNAV (GPS), RNAV (RNP) or GPS approaches.

Number of airports without RNAV approaches by state
TX with 7
OH with 4
CA and SC with 3
AZ and UT with 2
AK, AL, CT, FL, ID, IL, LA, MN, NV, OR, SD and WI with 1

List of airports:
12G; 1X1; 2P7; 44C; 4I9; 56D; BAD; C16; COF; DYS; F14; F41; HIF; LSV; LUF;
LZD; MCC; MGG; MMT; MUO; MXF; NOG; NQI; P19; PHH; RCA; S12; SSC; U69; VBG;
VDZ; VNY; WEA

On the opposite end, there are 1,406 airports with only RNAV (GPS) or GPS approaches.

That seems like the definitive answer, however I don't understand where the data is coming from. Any chance you could provide a link to this raw data?

Downvote for what is almost certainly a too-narrow definition of "airport" used without providing the actual definition. For example, in my airspace alone there are three charted public-use (three-character LID) airports with no instrument procedures at all, and that's not even counting the fifteen-plus private-use airfields. And I am by no means exceptional; pull up Skyvector and see for yourself.

@randomhead that's perhaps an unfair downvote. If you read my question, I specifically ask about airports which have instrument procedures, but no RNAV ones. I'm assuming that wbeard52 was answering in this context of "airport".

The data is coming from the U.S. digital TERPS download.  https://www.faa.gov/air_traffic/flight_info/aeronav/digital_products/dtpp/
On the bottom of the page it has downloadable charts.  Group E has a XML that can be parsed.  That is where the data came from.  It is for airports with at least one instrument approach.  Not VFR airports.

@wbeard52 Could you provide the search query you used to parse the XML?  I think your answer is more accurate than Bianfable's, but I'd like to compare the two.

I downloaded the "E" version of the TERPS packet.  Pulled out the XML file from the ZIP.  I put imported the XML file in Excel.  I then created a formula to provide me a column with the type of approach for each row.   I then created a pivot table with the corresponding data.  Once I had the airports, i input the airports into a database where I have catalogued information from the FAA runway database to get the airport name, city and state.

I'm not finding an easy way to recreate the query without Excel, but perusing the d-TPP_Metafile.xml file downloaded from the DDTPPE_XXXXX link, I can see where the data originated from. It'd be great if someone could comment with a CLI query which we could run periodically to keep this post up to date.

0BB doesn't seem to be an airport or otherwise figure in the XML file. Is it possible that was a typo? The random selection of others check out, including Y47 which is a one of the handful of airports I've landed at in the last decade. Go figure...

I assume you’re limiting this dataset to public-use (three-letter/number) airports?

I don't really know about that.  Whatever is in the dataset was parsed.  I would assume it does not include special approaches or privately owned approaches that are FAA created.

@wbeard52 I made a flight plan in Skyvector to all those airports, and a surprising number have some kind of "VOR or GPS". For instance 17N, 29N. DLC, ETC, NSE, etc... I wish I could figure out a textual search so we could refine it as we discover edge cases.

@KennSebesta “GPS overlay” on conventional approaches used to be very common, but most overlays were replaced by distinct RNAV approaches over time. Apparently there’s still a few they just haven’t gotten around to yet.

I was revisiting the data and noticed the "or GPS" as well.  I went ahead and pulled the airports with no RNAV or GPS approaches.  I was curious to know how many airports only have GPS approaches as well.

Very interesting question! I analysed my current navigation database*1 (AIRAC 2205), filtered for US airports with instrument approaches and grouped them into 3 categories:

airports with radio based approaches, but without RNAV/RNP approaches
airports with RNAV/RNP approaches, but without radio based approaches
airports with both

The result is shown here:

Only 25 airports (1.8%) have a radio based approach, but no RNAV/RNP available. The list is here:
K2K7, KAVX, KBAD, KC03, KCOF,
KDYS, KHIF, KINW, KIPL, KLSV,
KLUF, KLZD, KMCC, KMIB, KMMT,
KMUO, KMXF, KNQI, KNRB, KOLS,
KRCA, KSOA, KSSC, KVBG, KVNY

How does this compare to the rest of the world?

As you can see, the percentage of airports without RNAV/RNP is much higher in the rest of the world. In particular, European*2 airports still rely much more on radio based approaches (28.6% have radio only). The global average is 18.2% for radio based only.

*1 My database does not contain all airports in the US, usually only the ones with instrument approaches available, but there is no guarantee for completeness.
*2 For European airports, I filtered for those airports with ICAO airport codes starting with E or L, which covers most of Europe.

Which airport is assigned the identification code K2K7, or KC03? I think you will find there are no such airports, in any identification scheme. (This is an error in your database...)

@randomhead K2K7 is Neodesha and KC03 is Nappanee Municipal Airport (the database uses ICAO identifiers, but these 2 don't have an ICAO identifier, so it uses K followed by the FAA code). Regarding your second point: The question says "with instrument approaches" in the body, just not in the title.

"K followed by the FAA code" does not an ICAO code make. Your database is incorrect; this is not an attack on you, I'm just pointing out that fact.

@randomhead You're right, that's not a correct ICAO code, but you'll have to complain to Jeppesen about that ;) I assume they are doing this because some aircraft avionics require entering the ICAO code for departure/destination, so they have to include some kind of workaround in the database.

Several listed airports there are military fields (KBAD, KDYS, others). Wonder if they have RNAV approaches, just not public ones. Two decades ago, the bases I flew out of had approved procedures for self-contained approaches, albeit probably not public & very possibly not via all FAA conventions.

This is a very interesting and effective presentation. @wbeard52's answer is more complete in the strict sense that it is exhaustive for US airports, with data pulled from the FAA itself. It'd be cool to merge those two answers...

I would be quite interested in a public AIRAC database cycle.  Where did you get your data from?  Is it public use?

@wbeard52 I don't think there is a (legal) free source for AIRAC data. I bought my data from Navigraph (for flight simulation), and they get the data from Jeppesen.

@KennSebesta - The answers actually seem to disagree which airports to list... i.e. one is not a subset of the other.  Which database is more trustworthy?

I tweaked the display of the list to make it more readable. 5 spaces turns it into a "code block" where the system will not flow text and will force side scroll.  Hard to read, near impossible on some mobile OSs.

@IronEagle I think for the purposes of this question, where there is no safety factor, we have to take the FAA database as the canonical truth. I do find it interesting that there is so much disagreement in the list, but that the general result-- there are very few procedured airports without RNAV approaches-- is the same. I think once we hit upon a textual XML search we'll be able to figure out the truth of the matter.

Yes, there are airports without RNAV approaches. They are very common; as another answer pointed out, there are fewer than 7000 RNAV approaches spread over 19,400 airports in the NAS, and they are certainly not spread evenly. If you browse Skyvector you will see any number of examples without even trying.
As far as larger public-use airports, KMCC (McClellan Airfield, Sacramento, CA) is an example. At the time of writing, it only has ILS and VOR/DME approaches. Others nearby include KVNY (Van Nuys, CA) and KLSV (Nellis AFB).
Why? Presumably they'll get around to it eventually. Designing an approach isn't "trivially easy". It takes considerable work on the part of the FAA to design a new approach.
(Just as a reference: As of 2016, an extremely basic RNAV approach with one line of minima starts at $10,000 - not including annual maintenance costs. Which may be cheaper than whatever it's replacing... but just to show that there's non-trivial work involved.)

Compared to the process of installing and maintaining a VOR, the considerable work is considerably small. One might even say it's trivial!

Last year a city near me removed their ILS (and the only BC within 200 miles) because it cost them about $1M/yr to maintain. $10k for an RNAV approach is arguably trivial in comparison.

To be clear, my point is not that an RNAV approach wouldn't be cheaper than whatever's already present. Just that it takes time and effort to create a new approach.

@TrevorJohns Santa Monica has had RNAV approaches for several years now: https://nbaa.org/aircraft-operations/airports/smo/new-rnav-approaches-enable-greater-access-santa-monica-airport/

Not to over simplify things, but if all the TERPs work has been done for an existing ILS or VOR approach, how much more actually needs to be done?  Most GPS databases already contain ILS approaches that you can load to overlay and monitor the ILS, couldn't the FAA wave their magic wand and declare that the overlay is now an approved RNAV?  It seems like it would be a largely administrative drill...

@757toga That’s per the city aviation department, who funded their own ILS for decades after the FAA rejected them.  (The city’s hope was having an ILS would draw traffic from other airports that didn’t have one.)


text:
Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
We had a discussion with friend and we were talking about if the airplane is on approach and the one is landing with tailwind and the other one is landing with headwind, so both have the same value of IAS but the different value of ground speed. Will the airplanes produce the same value of lift in both cases or will be the value of lift be different? He said that value will be still the same and I was argueing that if you have tailwind you will have to increase angle of attack which means that parasite drag will increase so the value of lift will have to increase.
EDIT: Question is: If two same aircraft are on same glidepath (descending on ils glide slope) with same IAS but different ground speed, because of headwind acting on one aircraft and the tailwind acting on the other aircraft, will the amount of lift produced by one of the airplanes be different or will be both two values be the same?

Why would you have to increase the angle of attack? The air speed is the same, the air flowing over the wings is the same, the only difference is how long it takes you to stop once you touch the ground.

@RonBeyer, actually, there is a very subtle tricky bit! (assuming both aircraft follow the same glide-slope!)

Related -- https://aviation.stackexchange.com/q/87014/34686

EDIT: This question is really not about lift vectors. It is asking about the total amount or total value of lift needed with varying headwinds.
Everyone seems to be over complicating this very basic question which asks about two identical aircraft flying down an identical ILS glideslope with the exact same indicated airspeed. The only difference is the amount of headwind.
It is so simple. Rate of descent is a direct result of the amount of lift being produced.
It is a FACT that the aircraft with less headwind will need to produce LESS lift than the aircraft with more headwind, in order to maintain the same indicated airspeed, and follow the glide path.
The rate of descent with a headwind could be -500ft/min while the aircraft with a tailwind could be -1000ft/min.
The aircraft maintaining a -1000ft/min descent because of a tailwind will need to produce LESS lift than an aircraft maintaining a -500ft/min with a headwind.

Both aircraft will have exactly the same amount of total upward aerodynamic force, right? The only difference is that for the aircraft that's descending more quickly, relatively more of the upward force is drag and less of it is lift.

WHY has my answer been voted down? My answer is Correct. If you think it is not, please explain.

@MikeSowsun your answer describes the effects correctly. I can only assume that whoever downvoted it found it lacking an explanation for different sink rates requiring different amount of lift (and it may not be obvious for everyone - after all, descend is an unaccelerated flight state, so sum of forces must be zero, so why would you not need same lift for same weight?) :)

Mike has said "produce more lift" as in throttle.  See how we are somewhere between glider and level with our "produced" lift?  And how it affects our rate of descent at constant speed?

In this answer we read "Rate of descent is a direct result of the amount of lift being produced." That sounds like an Aristotelian point of view.  I.e. more upward force = faster upward velocity; less upward force = lower upward velocity or higher downward velocity. If that was not what was intended the answer could be clarified, for example by pointing out that the reduction in lift associated with a higher sink rate is real but extremely small, for reasonable ratios of L to (Drag minus Thrust). Also by pointing out that climbing also involves a reduction in Lift, compared to horizntl flight.

@ymb1 -- in the glide, the spoilers are functioning to increase drag, not to reduce lift.  The reduction in lift is a function of the relationship lift = weight* cosine (glide angle), and that will happen regardless of whether you make the glide slope steeper by deploying spoilers, or by deploying flaps and shoving the nose way down, or by pitching up to a very high angle-of-attack (possibly with flaps deployed) to descend "on the back side of the power curve", or by leaving the shape of the wing and the a-o-a of the wing the same and simply reducing power-- (ctd)

@ymb1 -- (ctd) The relationship lift= weight *cos (glide angle) still applies.  Just as when you configure the wing for the most efficient climb configuration, and add power to climb, lift is less than in level flight, regardless of a-o-a or wing shape.  Because lift = weight * cosine (climb angle).  But, I'm kind of just repeating my Oct 24  2018 comment I guess--

@ymb1 -- much more here now-- https://chat.stackexchange.com/rooms/123909/a-room-to-discus-question-descending-on-a-given-glide-slope-e-g-ils-at-a-give

A: The lifts produced by the two aircraft must be the same.
As given the vertical and horizontal speeds are constant. This requires all accelerations, including vertical acceleration, to be zero. Therefore the vertical component of lift must equal weight. This statement applies the same to both airplanes in their particular scenarios.
However knowledge that the vertical component of lift is equal does not make it obvious that the total lift is also equal. One argument is that AOA indicates the relationship between the vertical component of lift and total lift. Since both airplanes are straight path and same airspeed they have the same AOA. Because both airplanes have the same AOA and same vertical component of lift their total lift is also equal.

@ymb1 the acceleration I'd zero, hence upward force must equal weight. But the upward force in this case is composed partly of drag. It seems the author of this answer missed that.

In this answer we'll start by making the simplifying assumption that the Thrust vector is parallel to the Drag vector and the direction of the flight path with respect to the surrounding airmass.
See the vector triangles of Weight, Lift, and Drag for gliding flight illustrated in these related answers (#1, #2).  (The diagram in the second link is now reproduced at the top of the present answer as well.)
We can adapt these questions to a powered descent by simply substituting a vector labelled "Drag minus Thrust" in place of the vector labelled "Drag".  We'll still have a closed vector triangle.
Once we make this simple substitution, the angle labelled "K" in the diagram at the top of this answer will be the aircraft's angle of descent at some particular angle-of-attack and L/D ratio and Thrust output, in still air.  This is also the aircraft's angle of descent with respect to the surrounding airmass.
Note that Lift = Weight * cosine (K).
This means that Lift is less than Weight, and the steeper the still-air glide angle "K", the larger the difference between Lift and Weight.
To achieve the same descent path with respect to the ground with a headwind, we must change to a flatter descent path with respect to the surrounding airmass.  Whether we increase Thrust, decrease Drag, or both, we must make the vector we've re-labelled "Drag minus Thrust" become smaller.  This will decrease the size of angle K.  This will increase the size of the Lift vector.
To achieve the same descent path with respect to the ground with a tailwind, we must change to a steeper descent path with respect to the surrounding airmass.  Whether we decrease Thrust, increase Drag, or both, we must make the vector we've re-labelled "Drag minus Thrust" become larger.  This will increase the size of angle K.  This will decrease the size of the Lift vector.
Since Lift = Weight * cosine (K), the Lift vector must be larger when we are descending with a headwind than when we are descending with a tailwind, if we are achieving the same glide path with respect to the ground.
Note that this is true regardless of whether we're constraining the angle-of-attack to be the same in both cases, or the airspeed to be the same in both cases, or neither.  This is also true regardless of whether we're descending on the "front side" of the power curve or the "back side" of the power curve. None of these things modify the fact that to stay on a fixed glide slope relative to the ground, angle "K" must be larger when we are descending with a tailwind than when we are descending with a headwind.  Therefore the Lift vector must be larger when we are descending with a headwind than when we are descending with a tailwind.
Of course, as we vary angle K, the variation in the size of the Lift vector will be so small as to be negligible in the real world.  For example, when gliding with a still-air glide ratio of 7:1, which corresponds to a still-air glide angle of about 8 degrees, the Lift vector is about .99 * Weight.  If we were to change our glide path with respect to the airmass from a 7:1 glide ratio to a glide ratio of infinity (i.e. horizontal flight), the size of the Lift vector would only increase by a factor of about 1.01.  Still, technically speaking, the flatter our glide angle with respect to the surrounding airmass (i.e. the larger our still-air glide ratio), the larger the Lift vector must be.
So the basic answer to the question is that the Lift vector is ever-so-slightly smaller when the aircraft is descending with a tailwind than when the aircraft is descending with a headwind, because to stay on the same glide slope with respect to the ground, the still-air glide angle must be larger (i.e. the still-air glide ratio must be poorer) in the former case.
Let's consider one specific idea in the original question--

I was arguing that if you have tailwind you will have to increase
angle of attack which means that parasite drag will increase so the
value of lift will have to increase.

This is a flawed thesis.  To fly with the same airspeed, why would you need a higher angle-of-attack when you have a tailwind?  We've seen that-- disregarding any effects from a tilted thrust line relative to the direction of the flight path-- the Lift vector actually must be slightly smaller when the still-air glide path is steeper, which describes the tailwind case.  To keep the airspeed constant, the aircraft must be flown at a slightly lower angle-of-attack when descending with a steeper still-air glide path (i.e the tailwind case) than when descending with a shallower still-air glide path (i.e. the headwind case).
Now for another twist--
If we complicate the picture by recognizing that the Thrust line may be tilted to point somewhat upward or downward of the direction of the Drag vector and the flight path through the airmass, this may dwarf the effects described above.  The original question constrains the airspeed to be the same in the headwind case as the tailwind case, so the aircraft must be flying at nearly the same angle-of-attack in each case.   Imagine a case where we are flying our approach "on the back side of the power curve", at a high angle-of-attack, with the engine pointed significantly upward in relation to the direction of the flight path through the airmass.  The component of the engine's Thrust vector that acts perpendicular to the flight path effectively counts as part of the "Lift" vector in our vector diagram, meaning that the wing's actual Lift vector must be reduced to smaller than Weight * cosine (K).  In a case like this, reducing the power setting to make the glide path steeper may well force the wing's lift to increase.  On the other hand, if we are flying our approach at a low angle-of-attack, well on the "front side" of the power curve as per normal practice in a light airplane, it is no longer so obvious which direction the Thrust vector is pointing in relation to the direction of the flight path through the airmass.  Since many light aircraft have some built-in down thrust, the Thrust vector may well point slightly downward in relation to the direction of the flight path, so that the wing must produce more Lift when the engine's power setting is increased.  This would magnify the effect we discussed earlier in this answer-- the tendency for the Lift vector to be larger when the still-air glide angle is flatter, due to the relationship Lift = Weight * cosine (K).
At the end of the day, all these effects are so small that it's unlikely a pilot could detect the change in the magnitude of the wing's Lift vector for a given airspeed at various power settings and still-air glide angles.  But in theory there is some variation in the size of the Lift vector as the power setting and still-air glide angle are changed.  And for the simple case where we are assuming that the thrust line acts parallel to the Drag vector, the nature of the variation is such that the Lift vector is slightly larger when the still-air glide path is flatter (i.e. at higher power settings, such as when trying to stay on glide slope with a headwind) than when the still-air glide path is steeper (i.e. at lower power settings, such as when trying to stay on glide slope with a tailwind).
See also these related answers to related questions:
Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag? -- a concise answer with clear diagrams
Can we show through simple geometry rather than formulae or graphs that the best glide ratio occurs at the maximum ratio of Lift to Drag? -- a longer answer that emphasizes many concepts essential to the present answer
Why is the L/D ratio numerically equal to the glide ratio? -- similar to link immediately above
'Gravitational' power vs. engine power -- the variation in the size of the Lift vector as the still-air glide ratio changes comes into play here, exactly as it does in the present answer
Does lift equal weight in a climb?

You didn't answer the question: "will the amount of lift produce by one of the airplanes different or will be both two valeus the same?"

Yes, the question is answered.

I am tempted to downvote this answer merely on the grounds of ineffectiveness of getting any points across, regardless of accuracy. Long form links, no diagrams, relying on links to other answers to make points. However, what actually earns my downvote is that you ignore that the thrust vector will not generally not be aligned with the relative wind. The necessary variation in thrust to vary descent rate will therefore add to (rarely subtract from?) "K". Therefore you cannot make the assumption to simplify to (drag-thrust) as you do. Hence my downvote.

By long form links I refer to your practice here of inserting the entire hyperlink as plain text, rather than as an embedded link. Using "the answer https://aviation.stackexchange.com/a/56040/34686" instead of "this answer".

"K" was the variable used in one of your linked answers to name the vertical component of lift. I should have been more specific since it has no meaning outside of that context.

Let us continue this discussion in chat.

Re "However, what actually earns my downvote is that you ignore that the thrust vector will not generally not be aligned with the relative wind" -- the answer has now been reworked to address this point, and other issues.

Re ""K" was the variable used in one of your linked answers to name the vertical component of lift. I should have been more specific since it has no meaning outside of that context" -- unfortunately K also got used in one of my diagrams for something else; I won't take time to change it right now.

Related -- https://aviation.stackexchange.com/a/87040/34686

If both aircraft maintain a constant rate of descent, both aircraft will be generating the same amount of lift, as in both cases the motion is steady state ie not accelerating and therefore lift will be equal to gravity.  The aircraft descending with a tailwind will be doing so at a greater rate of descent, due to the higher ground speed and will require less thrust to maintain this state than the airplane landing in a headwind or still air, as more of the the potential energy can be converted into kinetic energy in the descent than in the case of the airplane descending with a headwind down an identical glideslope.

Lift is, by definition, the component of aerodynamic force perpendicular to relative wind. Since the descent rates, and thus the relative wind angles, will be different, the magnitudes of lift will be different for the vertical component to be the same.

The descent rate, in both cases, in relation to the air mass is the same. The descent rate in relation to the ground may be different, but relative wind angles are obviously determined by the motion in relation to the airmass, not to the ground...

Jan Hudec no, both airplanes, being equal weights, are in constant rate descents, therefore lift, by definition, MUST be equal in both cases.  However the aircraft descending with a tailwind would need to use less power to maintain stabilized flight and descent.

It sounds like you may be assuming that airplanes are held up entirely by lift. They're not; if an airplane is descending (relative to the air), then some of the force holding it up is lift, and some of the force holding it up is drag.

No, in this case lift is the sum of all forces acting perpendicular to the path of flight, and opposing gravity.   If lift was less than the force of gravity in this case, the airplane would begin to descend faster and faster until an equilibrium was again reached.

Again look at the vector diagram of an airplane descending.

I'm looking at the vector diagram of an airplane descending (I assume you mean the one here: https://aviation.stackexchange.com/questions/56023/gravitational-power-vs-engine-power). What I see on the diagram is that K (total aerodynamic force) is equal in length to W (weight), but L (lift) is shorter than K, so the diagram is indicating that lift is less than weight. Do you see something about the diagram that disagrees with what I just said? Also, I think that, since the drag force is not horizontal, drag contributes to holding the airplane up; do you disagree?

Tanner, what you are looking at is a glider descending at a speed lower than Vbg. If you pitch forward and increase airspeed, total lift vector increases enough so All vertical lift comes from wing.  I really hope this did not throw Jan off, as aircraft coming in too slow are asking for a stall, especially in wind, IMHO, an aircraft approaches with a combination of glide and powered flight, Not parachute!  Although one can come in near stall/sink, the very changes in relative wind Jan speaks of are a recipe for disaster (see glider crash question).

Actually there is, but I will not stress it.  With further thinking I have realized it can be rationalized as a combination of the vertical component of drag added to the vertical lift component.  What this does is frees the horizontal component of lift to react with gravity to produce a thrust vector on the line of flight.  This is good vector "bookkeeping" that I did not initially understand.  Apologies to Jan as well.  Amazing how we talk about the same things in different languages.  Working from the straight down case helped.


text:
Why does power go up with the cube of the airspeed?
If i have an airplane with over 100 kN of thrust, and i want to accelerate from a velocity at which drag force is 25kN to a velocity twice higher, is it possible that i wont be able to do this because my engine doesn't have enough power, even though it has enough thrust to counter the drag force near this higher speed, so there should be a net forward acceleration on my plane?

Related: https://aviation.stackexchange.com/questions/81458/why-is-thrust-said-to-be-constant-over-speed-for-a-jet-engine?noredirect=1&lq=1 ,https://aviation.stackexchange.com/questions/88618/why-is-thrust-inverse-to-speed-in-piston-engines? , https://aviation.stackexchange.com/questions/49946/how-and-why-does-engine-thrust-change-with-airspeed , https://aviation.stackexchange.com/questions/81809/how-do-power-and-thrust-curves-compare , https://aviation.stackexchange.com/questions/81857/what%c2%b4s-the-logic-behind-the-power-curve? , etc...

W = F*d
Where W is the energy needed to apply a force F over a given distance d.
Power is energy divided by time and also equivalent to Force applied for a given velocity.
F x v = F x d/t = W/t
For an airplane and straight and level flight at constant true airspeed, the force of drag is equal to the force of thrust.
Fd = Ft
Since the force of drag is proportional to the square of the velocity by
Fd = 0.5 * CdpA*v^2
Where p is equal to the density of the air, Cd is a coefficient of drag, A is equal to the flat plate area of the object being pushed through the air, and v is the true airspeed of the air, this gives the power consumed by drag applied at a given airspeed as
Pd = Pt = Fdv = 0.5CdpA*v^3
Pt is proportional to the cube of v, where Pt is engine power.

This answer assumes a constant drag coefficient-- shouldn't that be justified somehow, or at least noted?  In the context of flight of any given actual aircraft, where a-o-a must vary with airspeed?

Regardless of whatever changes the drag coefficient has, The force of drag still varies by the square of the airspeed, and thus the power required will vary by the cube of the airspeed

does this imply that an engine of given max power will lose thrust with increasing velocity?

Firstly, your hypothetical situation ("is it possible...") would represent an over-constrained problem. It would never be possible that an a/c has enough thrust to overcome drag at speed Y, but doesn't have enough power to overcome drag at the same speed Y. That's a fundamental contradiction, indicating a misunderstanding about the basic relationship between force, work, and power, possibly better suited to exploring on an engineering or physics site.
(This is assuming that by "has enough thrust" and "has enough" power, we mean at whatever particular airspeed we are talking about-- not the max thrust or power that we'd get at the optimum airspeed for maximizing that particular parameter.  It's not completely clear from the question which you meant-- if you meant the latter case, then we should note that it certainly does often happen that there are airspeeds we can't reach (in level flight) due to "lack of thrust" even if though we can produce that same amount of thrust at some lower airspeed, particularly with piston engines that tend to  have a very roughly constant power output, and thus experience a dramatic loss of thrust as airspeed is increased.)
Second, you haven't given enough us information to know or even guess the drag force at your second velocity Y(=2X).  (Obviously I'm calling your second velocity "Y" and your first velocity "X".)
Third, you haven't told us whether the 100 kN thrust force is constant independent of airspeed, or what.
The answer to the title itself, "Why does power go up with the cube of the airspeed?", is "It doesn't-- power required is not directly proportional to airspeed cubed across the entire flight envelope, because the drag coefficient is not constant".  Of course this is assuming that by "power" in the title, you meant the power required, not the maximum power that the propulsion system could produce at that airspeed.  The latter quantity, naturally, also is not directly proportional to airspeed cubed!
There are situations where an aircraft has enough thrust and power to overcome drag in horizontal flight at some given airspeed, but doesn't have enough thrust and power to overcome drag in horizontal flight at some lower airspeed, because that lower airspeed is on the "back side of the thrust-required curve" where drag is very high, so the thrust and power required (for horizontal flight) are also very high.  In such a case the aircraft could not accelerate, without giving up altitude, from the lower airspeed to the higher airspeed.  Nor could the aircraft maintain altitude at the lower airspeed.  But that doesn't seem to be what you are asking about here, as evidenced by your comment "so there should be a net forward acceleration on my plane".
To better understand how the power required for horizontal flight varies with airspeed, in any given actual aircraft (assuming fixed constant weight), where the drag coefficient is not constant because angle-of-attack must vary as airspeed varies, see these sections from John Denker's excellent See How It Flies website-- the graphs included here are much better than any verbal description, and you'll see that the required power does not simply vary according to the cube of airspeed--
Drag and the power curve-- introduction
More about the power curve
Related ASE questions that deal with how the thrust and power available from the propulsion system vary with airspeed
Why is thrust said to be constant over speed for a jet engine?
Why is thrust inverse to speed in piston engines?
How (and why) does engine thrust change with airspeed?
How do power and thrust curves compare?
What´s the logic behind the Power Curve??

"it would never be possible that an a/c has enough thrust to overcome drag at speed Y, but doesn't have enough power to overcome drag at the same speed Y" this is esentially the answer i wanted

@FrancisL.  -- glad I could help, sorry for so many comments, now have deleted most and incorporated into this answer!

@MichaelHall -I was only trying to say that if you have enough thrust to maintain a given airspeed, then you also have enough power, and if you have enough power to maintain a given airspeed, then you also have enough thrust. Assuming that we are talking about the thrust and power available at the speed in question, not the max possible thrust or power that would be available if the speed of the a/c were optimized for whichever of these two parameters we are talking about the moment. If we mean the latter, then it starts getting more complicated to explain what we are really trying to say.

@MIchaelHall -- have read it; honestly at the end of the day these kinds of questions (and answers) are much better served by graphs than by verbal descriptions!  Had serious second thoughts about answering the question at all due to multiple apparent misconceptions but-- since there was already another answer decided to go ahead--

@FrancisL. -- hope the links I've recently added to this answer to the See How It Flies website , and to related ASE answers, are helpful--

vary with airspeed--

$$F=m.a$$
At constant mass, we only need to consider the difference in thrust and drag forces to determine if the aircraft can accelerate to a higher velocity.

if i have an airplane with over 100 kN of thrust, and i want to accelerate from a velocity at which drag force is 25kN to a velocity twice higher

For incompressible flow, at standard aviation Reynolds number, drag force D is $$ D = C_D. \frac{1}{2} \rho V^2. S$$
At constant altitude and attitude, $D = C. V^2$ with C being a constant, so at twice the speed the drag is 4 times as high. Thrust T must be equal to this higher D, power does not appear in this consideration.
During engine design, required engine power is determined by required aerodynamic power = $T. V$ divided by transmission and thrust mechanism efficiencies. It is the aerodynamc required power that goes up by the cube of velocity, engine power must be able to handle that.

Same comment I left the other answer -- This answer assumes a constant drag coefficient-- shouldn't that be justified somehow, or at least noted? In the context of flight of any given actual aircraft, where a-o-a must vary with airspeed?

@quietflyer “At constant altitude and attitude..”


text:
Trying to find the vertical acceleration by subtracting weight from lift
I'm trying to determine the vertical acceleration component of a plane with the following specifications:

Density of air: 1.29kg/m3
Speed: 33 m/s
Wing area: 13.97 m2
Lift Coefficient: 1.5
Mass:560 kg

Lift formula : 1/2*1.29*33^2*13.97*1.5 = 14718 N.
(Lift-Weight)/Mass = (14718-(560*9.8))/560 = 16.48
From this I deduce that the plane has a vertical acceleration of 16.48 m/s2 which I find way too much.(I was expecting to get somewhere from 3 to 5 m/s2)

your mass seems a little low and your speed and wing area a little high.  Your lift coefficient is generous.  So what combination of pilot, fuel, airframe, and engine gave you 560 kg?  But you are close enough to a Cessna 152 to compare real data to your calculations.  Also, remember increasing Lift coefficient to 1.5 on a wing that big is going to have a fairly big drag penalty, requiring more thrust, requiring a larger engine, which weighs more ... etc.  Check that 152 out for starters.

PS crop dusters have lots of power for just that type of performance, you may want to check them too.

The lift grows with square of velocity and note that quadratic curve gets steep rather quickly. At typical approach speed of 1.3 times the stall speed, you can have vertical acceleration of 6.56 m/s² already and at twice the stall speed you can have almost 30 m/s². 3 to 5 m/s² would only be dangerously close to the stall speed. It's only available vertical acceleration of course—most of the time you simply don't have any, flying at lower angle of attack to just balance your weight.

Your math looks right to me. If an airplane has the statistics that you listed there, then it will accelerate upwards at $16.48\ \mathrm{m}/\mathrm{s}^2$, as you calculated. The occupants will experience about 3 g's of proper acceleration (1 g from gravity, and about 2 g's from the acceleration).
But here's what would actually happen, in context.
To start, assume that the plane is in straight and level flight at 33 m/s. (Here's why I'm making this assumption: You're subtracting weight from lift, which only makes sense if the lift vector is pointing straight up. And if the lift vector is pointing straight up, then the plane is probably in straight and level flight.)
While the plane is in straight and level flight, the lift must equal the weight, so the coefficient of lift must be much less than the 1.5 you have here.
Then, the pilot suddenly pulls back on the yoke, causing the plane to pitch up. This causes the angle of attack to quickly increase, which causes the coefficient of lift to increase to 1.5. The plane starts accelerating upwards at $16.48\ \mathrm{m}/\mathrm{s}^2$.
At this point, let's assume that the pilot maintains a constant pitch after this initial pull-up.
As the plane's vertical speed increases, its climb angle also increases, which causes the angle of attack to decrease. This, in turn, causes the coefficient of lift to decrease again, until the lift and weight approximately balance out again. At this point, the plane is in a steady climb. The vertical acceleration is now $0\ \mathrm{m}/\mathrm{s}^2$, and the occupants are feeling just 1 g again.
(I'm making a couple of small simplifying assumptions in the above: I'm assuming that there is no updraft or downdraft, and that plane's engine is not producing a significant amount of upward force.)

I would say what the original question describes is something like the start of a loop.  The pilot's controI input that creates the 1.5 CL condition also initiates a pitch rotation.  I would not say that the resulting upward acceleration automatically decreases the plane's angle-of-attack, because the plane is free to rotate in pitch.  However two effects that do tend to decrease the angle-of-attack associated with any given elevator position are pitch rotational inertia (as the maneuver is initiated) and aerodynamic damping of pitch rotation rate (once a significant  pitch rotation rate is

However two effects that do tend to decrease the angle-of-attack associated with any given elevator position are pitch rotational inertia (as the maneuver is initiated) and aerodynamic damping of pitch rotation rate (once a significant  pitch rotation rate is achieved).  Despite these effects, to transition to a steady-state climb after initiating a 1.5 CL pullup, I think the pilot would probably have to relax his elevator input, at least until some of the airspeed bleeds off.

@quietflyer That's a good point. I've edited this to point out that I'm assuming that the pilot maintains a constant pitch after initially pulling up.

Comments now developed into an answer...

I would say what the original question describes is something like the start of a loop.  The pilot's elevator controI input that creates the 1.5 CL (about 3G in this case) condition also initiates a pitch rotation.  To a first approximation, I would not say that the resulting upward acceleration automatically decreases the plane's angle-of-attack, because the plane is essentially free to rotate in pitch to maintain the angle-of-attack that is commanded by the elevator position.  However, looking in more detail, two effects that do tend to decrease the angle-of-attack associated with any given elevator position (or any given pull force exerted on the stick or yoke) are pitch rotational inertia (as the maneuver is initiated) and aerodynamic damping of pitch rotation rate (once a significant  pitch rotation rate is achieved).  Despite these effects, to quickly transition to a steady-state climb after initiating a 1.5 CL (about 3G in this case) pullup, the pilot would have to relax his elevator input, at least until some of the airspeed bleeds off.
Keep in mind that in a steady-state climb, lift must be LESS than weight.


text:
Why are these characteristics of the FT-180 highlighted?
I have been checking the FT-180 engine used to establish speed record for remotely controlled aircraft (i.e. model airplanes). There are some specific features highlighted on the manufacturer site. I'm not specialist of turbojets, but I don't understand why those characteristics are highlighted:

They are at the top of innovation with unique features like carbon fiber cowling, Genx type chevron exhaust system and full Inconel internals.
===> Since all these materials are explicited, what is the specific element behind? How do they prevent another company to build a similar engine?
All aluminium and steel design: no plastic parts are present on the engine except for the carbon fiber inlet cowl.
===> Why is this important no to have plastic part? I thought that plastic was a light and weather resistant material, so a good one for a remotely controlled aircraft?
Genx type chevron exhaust cone .
===> Genuine question: why is this important? Does it give higher performance to the engine?
Also, with link to this question: This engine is a axial engine, not a centrifugal one, am I right?

This is a question about an engine for model airplanes-- I hope my edit made that clearer.

Thanks for the edit

It is marketing material. Marketing is about making your product (the interfiber multipurpose connector) different from the competitors (a nail).

Since all these materials are explicited, what is the specific element behind?

Carbon fiber is known for being light weight and inconel is known for high temperature capabilities. So by naming the materials they are claiming the engine is lightweight and high performance

How do they prevent another company to build a similar engine?

Not naming the materials would not prevent competition. Inconel is very common material for hot components in turbines. Anybody else building something with a similar price and performance will probably have to use inconel. There's not too many other options. Same with carbon fiber

All aluminium and steel design: no plastic parts are present on the
engine except for the carbon fiber inlet cowl. ===> Why is this
important no to have plastic part? I thought that plastic was a light
and weather resistant material, so a good one for a remotely
controlled aircraft?

Because plastics have a relatively low melting point, and jet engines get hot.